Physics · Moving Charges And Magnetism · NEET
n is turns PER UNIT LENGTH, meaning turns per metre: n = N/L. This is the single most common mistake. If a solenoid has N = 100 turns over L = 0.5 m, then n = 100/0.5 = 200 turns per metre, NOT 100. Always divide total turns by the length in metres before you multiply. Watch the units: if length is given in cm or mm, convert to metres first.
For a long solenoid (length much greater than radius), the field deep inside is uniform and depends only on how tightly the turns are packed (n) and the current (I), not on how wide the tube is. In NEET 2022 a solenoid of radius 1 mm was given - the radius was a pure distractor. B = mu0 n I ignores it. Only n and I matter for the interior field.
At the exact centre (interior) of a long solenoid the field is B = mu0 n I. At either open END, the field drops to exactly HALF: B_end = (mu0 n I)/2. Reason: the end sees only half of the winding contributing symmetrically. NEET can test this - if the question says 'at the end' or 'at the mouth', use B/2.
The fields from opposite sides of the winding cancel outside the tube, so for an ideal long solenoid the external field is negligibly small (nearly zero). We USE this in the Ampere's law derivation: only the path segment inside the solenoid contributes. This is why a solenoid acts like a bar magnet with the field confined inside.
No. The whole point of a long solenoid is a UNIFORM field. Anywhere in the interior (away from the ends), B = mu0 n I has the same value and direction, parallel to the axis. This uniformity is exactly why solenoids are used to make controlled magnetic fields in experiments and machines.
A long solenoid of 50 cm length having 100 turns carries a current 2.5 A. The magnetic field at the centre of the solenoid is (mu0 = 4pi x 10^-7 T m A^-1)
A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For the interior of a long solenoid, B = mu0 * n * I, where n = N/L is turns per metre, I is current, and mu0 = 4pi x 10^-7 T m/A. Outside it is nearly zero.
Apply Ampere's circuital law to a rectangular loop with one side of length L inside the solenoid and one outside. Field outside is zero and the side ends are perpendicular to B, so only the inside side contributes: B*L = mu0 * (n L) * I, which gives B = mu0 n I.
At either end of a long solenoid, the field is half the interior value: B_end = (mu0 n I)/2.
No. For a long solenoid the interior field depends only on n (turns per metre) and I. The radius does not enter the formula, which is a common NEET distractor.
It points along the axis of the solenoid. Use the right-hand rule: curl fingers along the current in the turns, and the thumb points along B (the north pole end).
